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From: Charlie-Boo on 20 Jan 2010 19:47
One obvious case is when proving by induction. Any others?

C-B
From: Jesse F. Hughes on 20 Jan 2010 19:53
Charlie-Boo <shymathguy(a)gmail.com> writes:

> One obvious case is when proving by induction. Any others?

Proving that every number divisible by four is also divisible by two
is easier than proving that

575759329310928935320580898273492384616575757346234296743565737100

is divisible by two.

Of course, there are a trillion similar examples.

--
Jesse F. Hughes
"Women aren't that unpredictable."
"Well, I can't guess what you're getting at, honey."
-- Hitchcock's _Rear Window_
From: scattered on 20 Jan 2010 21:16
On Jan 20, 7:47 pm, Charlie-Boo <shymath...(a)gmail.com> wrote:
> One obvious case is when proving by induction. Any others?
>
> C-B

One of the points of a good abstraction is that it allows you to prove
stronger results with less work. It is easier to prove that the
composition of continuous functions between topological spaces is
continuous than it is to prove the corresponding result for metric
spaces (a trivial proof in both cases but in the metric case you have
to trot out epsilons and deltas whereas it is virtually obvious in the
general case).

-scattered
From: Larry Hammick on 21 Jan 2010 00:49
"scattered"
On Jan 20, 7:47 pm, Charlie-Boo <shymath...(a)gmail.com> wrote:
> One obvious case is when proving by induction. Any others?
>
> C-B
[
One of the points of a good abstraction is that it allows you to prove
stronger results with less work. It is easier to prove that the
composition of continuous functions between topological spaces is
continuous than it is to prove the corresponding result for metric
spaces (a trivial proof in both cases but in the metric case you have
to trot out epsilons and deltas whereas it is virtually obvious in the
general case).

-scattered
]

Good point. It can be good play to weaken the hypotheses early in the game,
if possible. Along those lines, I'm right now writing up a conjecture that
is considerably stronger than the four colour theorem, but technically
simper to state, because it applies to some graphs that are not planar as
well as all graphs that are planar. There are a couple of combinatorics
experts near me on whom I will be trying it out.

LH


From: Bill Taylor on 21 Jan 2010 01:12
"Larry Hammick" <larryhamm...(a)telus.net> wrote:

> Along those lines, I'm right now writing up a conjecture that
> is considerably stronger than the four colour theorem, but technically
> simper to state, because it applies to some graphs that are not planar as
> well as all graphs that are planar.

May we be permitted an advance preview of the statement of the
conjecture?

-- Wondering William

** Postmodernists still fly in aeroplanes rather than on broomsticks,
** even though they "think all views are equally valid".
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